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I want to prove the following:

Assume we have two space groups $G,G^\prime \subseteq \text{Euc}(V) \subseteq \text{Aff}(V)$ which are affinely equivalent, $G \sim G^\prime, \; \text{ i.e. }\; \exists\, (\xi,a)\in \text{Aff}(V)\,:\; (\xi,a)\,G\,(\xi,a)^{-1} = G^\prime $.

Now define

$T(G) := G \cap \{(\tau,\mathbb{1}\;|\; \tau\in V)\}$ the normal subgroup and $P(G) := \{a\in O(V) \;|\; (\tau,a)\in G \text{ for some $\tau \in V$} \}$ the point group of $G$

(correspondingly for $G^\prime$).

The claim is that if the short exact sequence

$1 \longrightarrow T(G) \longrightarrow G \longrightarrow P(G) \longrightarrow 1$

splits, then so does the corresponding sequence for $G^\prime$.

I have already proven, that for the commuting diagram with short exact rows

$1 \longrightarrow G_1 \longrightarrow G_2 \longrightarrow G_3 \longrightarrow 1$

$\quad \quad \;\;\:\downarrow \quad \quad \;\;\;\downarrow \quad \quad \;\;\downarrow$

$1 \longrightarrow H_1 \longrightarrow H_2 \longrightarrow H_3 \longrightarrow 1$

with groups $G_i,H_i$ and homomorphisms

$\varphi_i: \; G_i \rightarrow G_{i+1}$,

$\psi_i: \; H_i \rightarrow H_{i+1}$,

$f_i: G_i \rightarrow H_{i}$,

it holds that

$f_1,f_3 $ injective $\Rightarrow\; f_2$ injective,

$f_1,f_3 $ surjective $\Rightarrow\; f_2$ surjective.

I'm not sure how to proceed here. I think that if I find isomorphisms $T(G)\rightarrow T(G^\prime)$ and $P(G) \rightarrow P(G^\prime)$, I'm done, since then I can find all needed inverses easily from the (assumed) inverses of the sequence for $G$. I'm also not sure how I could possibly use what I have proven already, because the isomorphism from $G$ to $G^\prime$ is just conjucation (or am I mistaken here?). However, I'm not a specialist on homological algebra, but merely a physics undergrad.

Any ideas?


I have thought about it a bit more, and have come to the conclusion that it suffices to find an isomorphism $\beta\,:\;P(G)\longrightarrow P(G^\prime)$. Let's give a name, i.e. $\theta\,:\; G\longrightarrow P(G)$. The sequence for $G$ splits, iff there exists $u\,:\;P(G)\longrightarrow G$ with $u\circ \theta = \mathbb{1}_G$ (by definition, right?). So if we assume $u$ splits the sequence for $G$, and we find two isomorphisms $\alpha\,:\;G\longrightarrow G^\prime$ and $\beta\,:\; P(G)\longrightarrow P(G^\prime)$, then $\alpha\circ u \circ \beta^{-1}\,:\;P(G^\prime) \longrightarrow G^\prime$ splits the sequence for $G^\prime$. Is this correct?

The isomorphism $\alpha$ exists by assumption, let's say $(g,a)\mapsto (\xi,b)(g,a)(\xi,b)^{-1}$. For the isomorphism $\beta$, I found $a \mapsto bab^{-1}$, where $b$ is the same as in $\alpha$. Then $\beta^{-1}(q) = b^{-1}qb$, and if all this is true i'm done. Looks good to me, what do you think? The remaining problem is now proving that the resulting map indeed splits the $G^\prime$ sequence, and I have no idea how to go about this, because nothing about the map $G^\prime \longrightarrow P(G^\prime)$ is known. I feel that I should prove that the diagram is commutative, or do I get this for free from somewhere?

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I have edited my old answer to fill it out into a complete answer. – Lee Mosher Jul 14 '14 at 13:52
up vote 3 down vote accepted

Are you making use of the conjugation $(\xi,a)\,G\,(\xi,a)^{-1} = G^\prime$? That's what gives the isomorphism from $G$ to $G'$, and it would be a good guess that this isomorphism takes $T(G)$ to $T(G')$ and induces an isomorphism of quotient groups $P(G)=G/T(G) \to P(G')=G'/T(G')$.

Edited much later:

They key fact needed is that the translation subgroup $T(V) \subset Aff(V)$ is a normal subgroup of $Aff(V)$. So if $A : V \to V$ is an affine transformation and $AGA^{-1}=G'$ then $$A \, T(G) \, A^{-1} = A \,(G \cap T(V)) \, A^{-1} = (A G A^{-1}) \cap (A \, T(V) \, A^{-1}) = G' \cap T(V) = T(G') $$ Let me use $i_A : G \to G'$ to denote the isomorphism defined by conjugating with $A$, that is, $i_A(g) \mapsto AgA^{-1}$.

Lemma: If $i:G \to G'$ is an isomorphism, if $N \subset G$ is normal, and if $N'=i(N) \subset G'$ is normal, then $i$ induces an isomorphim $i_* : G/N \to G'/N'$. This is an immediate consequence of applying the First Isomorphism Theorem to the composed homomorphism $G \xrightarrow{i} G' \mapsto G'/N'$ whose kernel is clearly $N$.

Lemma: A short exact sequence $1 \to N \to G \to Q \to 1$ splits if and only if there exists a retracting homomorphism $r:G \to G$ onto some subgroup of $G$ such that $N = Ker(r)$. Both directions of the "if and only if" are consequences of applying the First Isomorphism Theorem. For the "if" direction one composes the isomorphisms $Q \approx G/N \approx image(r)$ to get the splitting map, and for the "only if" direction, given a splitting map $s:Q\to G$, one composes the maps $G \to Q \xrightarrow{s} G$ to get the retraction.

(By the way, the statement of the First Isomorphism Theorem on that Wikipedia link is rather lazy. The theorem gives you more than just the abstract existence of an isomorphism. By tracing through the proof you get a very particular isomorphism, namely the induced map of the given homomorphism. This is needed in the above applications of the First Isomorphism Theorem).

So, putting this all together, suppose that $G'$ splits, so there is a retracting homomorphism $G' \xrightarrow{r'} G'$ whose kernel is $T(G')$ ($image(r')$ is of course equal to the image of the splitting $P(G') \to G'$, but we don't need to use that fact at this stage). Define a homomorphism of $G$ as follows: $$r: G \xrightarrow{i_A} G' \xrightarrow{r'} G' \xrightarrow{i_A^{-1}} G $$ Tracing through this, one sees that $r$ is a retraction, and using that $T(G') = Ker(r')$ and $i_A(T(G))=T(G')$ one sees that $T(G)=Ker(r)$. Therefore $G$ splits.

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From a quick calculation I did, it looks like you are right, conjugation seems to be an isomorphism $T(G)\rightarrow T(G^\prime)$ aswell. Now I'm stuck at the point of finding the induced isomorphisom between the quotients, any idea how this could work? – Pascal Engeler Jun 9 '14 at 19:52
Just use one of the standard theorems of group theory. If you have an isomorphism between two groups $f:G \to G'$, and if the isomorphism takes a normal subgroup $N$ to a normal subgroup $N'$, then the induced map $G/N \to G'/N'$ that takes $gN$ to $f(g)N'$ is a well-defined isomorphism. – Lee Mosher Jun 10 '14 at 16:01
is it essential that the homomorphism $r$ is a retraction? I don't see where that's used in the proof. I assume a retraction is a homomorphism $r:G\rightarrow H\leq G$ with the restriction of r on $H$ being the identity. Where is the last part of this needed? – Pascal Engeler Jul 18 '14 at 12:14
I don't exactly understand the proof of the second lemma. How do you verify that $Q\rightarrow G/N \rightarrow \text{Im } r \rightarrow Q = Id_Q$? Or is this clear because $\text{Im }r$ is isomorphic to $Q$? – Pascal Engeler Jul 18 '14 at 15:27
@PascalEngeler: The isomorphisms $G/N \to Q$ and $G/N \to \text{image}(r)$ are induced, respectively, by the surjection $G \to Q$, and the retraction $r$ (this is the kind of information that is missing from a lazy statement of the First Isomorphism Theorem). So if you define the splitting homomorphism by composing the inverse of the isomorphism $G/N \to Q$ followed by the isomorphism $G/N \to \text{Im}(r)$, it will follow with a bit of thought regarding cosets of $N$ that the composition you write down is indeed $\text{Id}_Q$. – Lee Mosher Jul 19 '14 at 15:49

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